Question

1. Let T-Σ-iz, where Z1,Zo, replacement from the set {1,2,... , N Show that ,Žm are numbers sampled at random without E(Zi) (N +1)/2 and hence E(T) m(N + 1)/2. Show that E(Z) 12 and hence - m)(N 12 Deduce that under the null hypothesis that F- G, the expectation and variance of Wilcoxon's two-sample test statistic are m(n+m+1)/2 and nm(n+m+1)/12, respectively.

Answer 1

$P(Z_i=j)=\frac{\binom{N-1}{m-1}}{\binom{N}{m}}=\frac{1}{N},~j=1,2,...,N\\ E(Z_i)=\sum_{j=1}^NjP(Z_i=j)=\frac{1}{N}\sum_{j=1}^Nj=\frac{N+1}{2}\\ E(T)=\sum_{i=1}^mE(Z_i)=\frac{m(N+1)}{2}.\\ E(Z_i^2)=\sum_{j=1}^Nj^2P(Z_i=j)=\frac{1}{N}\sum_{j=1}^Nj^2=\frac{(N+1)(2N+1)}{6}=\frac{2N^2+3N+1}{6}\\ P(Z_i=k,~Z_j=l)=\frac{\binom{N-2}{m-2}}{\binom{N}{m}}=\frac{1}{N(N-1)},~k\neq l=1,2,...,N\\ E(Z_iZ_j)=\sum_{k\neq l}^Nkl\frac{1}{N(N-1)}=\frac{1}{N(N-1)}\sum_{k=1}^Nk\left(\sum_{l=1}^Nl-k \right )\\ =\frac{1}{N(N-1)}\left(\left(\sum_{k=1}^Nk \right )\left(\sum_{l=1}^Nl \right )-\sum_{k=1}^Nk^2 \right )=\frac{1}{N(N-1)}\left(\frac{N^2(N+1)^2}{4}-\frac{N(N+1)(2N+1)}{6} \right )\\ =\frac{(3N^2+3N-4N-2)(N+1)}{12(N-1)}=\frac{(3N+2)(N-1)(N+1)}{12(N-1)}\\ =\frac{3N^2+5N+2}{12},~i\neq j\\ Cov(Z_i,Z_j)=E(Z_iZ_j)-E(Z_i)E(Z_j)=-\frac{N+1}{12}\\ ~Var(Z_i)=E(Z_i^2)-E^2(Z_i)=\frac{N^2-1}{12}$$Var(T)=\sum_{i=1}^mVar(Z_i)+\sum_{i\neq j}^mCov(Z_i,Z_j)\\ =\frac{m(N^2-1)}{12}+m(m-1)\left(-\frac{N+1}{12} \right )=\frac{m(N+1)(N-m)}{12}\\ For~Wilcoxon's~two~sample~test,~we~arrange~n~X's~and~m~Y's~in~increasing~order\\ test~statistic=W=sum~of~the~ranks~of~Y's~in~combined~sample.\\ Here,~N=n+m\\ Z_1,...,Z_m~are~ranks~of~Y's~in~combined~sample~so~W=T.\\ Hence~E(W)=E(T)=m(m+n+1)/12,~Var(W)=Var(T)=\frac{mn(n+m+1)}{12}$